An aircraft flying horizontally 100m above the ground and at 720km/h drops a bomb on a target on the ground.Determine the acute angle between the vertical and the line joining the aircraft and target at the instance when the bomb is released.(g=10ms^(−2) )


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Question Number 47333 by Necxx last updated on 08/Nov/18

$A n a i r c r a f t f l y i n g h o r i z o n t a l l y$ $100 m a b o v e t h e g r o u n d a n d a t 720 k m / h$ $d r o p s a b o m b o n a t a r g e t o n t h e$ $g r o u n d . D e t e r m i n e t h e a c u t e a n g l e$ $b e t w e e n t h e v e r t i c a l a n d t h e l i n e$ $j o i n i n g t h e a i r c r a f t a n d t a r g e t a t$ $t h e i n s t a n c e w h e n t h e b o m b i s$ $r e l e a s e d . (g = 10 {m s}^{- 2})$
	Answered by MrW3 last updated on 08/Nov/18

	$v = 720 k m / h = 200 m / s$ $x = v t$ $y = \frac{{g t}^{2}}{2} \Rightarrow t = \sqrt{\frac{2 y}{g}}$ $\Rightarrow x = v \sqrt{\frac{2 y}{g}}$ $\tan θ = \frac{x}{y} = v \sqrt{\frac{2}{g y}} = 200 \sqrt{\frac{2}{10 \times 100}} = \frac{20}{\sqrt{5}}$ $\Rightarrow θ = \tan^{- 1} \frac{20}{\sqrt{5}} = 83.6 °$
	Commented by Necxx last updated on 08/Nov/18

	$t h a n k y o u s i r . T h i s w a s w h a t I a l s o$ $g o t b u t t h e a n s w e r s a y s 70.5 i n$ $t h e t e x t$
	Commented by MrW3 last updated on 08/Nov/18

	$t h e n t h e a n s w e r i n b o o k i s w r o n g .$
	Answered by ajfour last updated on 08/Nov/18

	$h = \frac{1}{2} {g t}^{2}$ $D = v t$ $h = \frac{g}{2} {(\frac{D}{v})}^{2} \Rightarrow D = v \sqrt{\frac{2 h}{g}}$ $\tan θ = \frac{D}{h} = \frac{v \sqrt{2}}{\sqrt{g h}} = \frac{200 \sqrt{2}}{\sqrt{1000}} = 4 \sqrt{5}$ $θ = \tan^{- 1} (4 \sqrt{5}) .$
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